Page 21 - Schaum's Outline of Differential Equations

4 BASIC CONCEPTS [CHAR 1

1.4. Is y(x) = 1 a solution of /' + 2/ + y = xl
From y(x) = 1 it follows that y'(x) = 0 and ;y"(;e) = 0. Substituting these values into the differential equation,
we obtain

Thus, y(x) = 1 is not a solution.

1.5. Show that y = In x is a solution of ;ry" + / = 0 on J> = (0, °°) but is not a solution on J> = (- °°, °°).
2
On (0, °°) we have ;/=!/* and /' = -I/* . Substituting these values into the differential equation
we obtain

Thus, y = In x is a solution on (0, °°).
Note that y = In x could not be a solution on (- °°, °°), since the logarithm is undefined for negative numbers
and zero.

2
2
1.6. Show that y = l^x - 1) is a solution of y' + 2xy = 0 on J> = (—1, 1) but not on any larger interval
containing J>.
2
On (-1, 1), y= ll(x — 1) and its derivative y' = — 2xl(x 2 — I) 2 are well-defined functions. Substituting these
values into the differential equation, we have

Thus, y = l/(x 2 - 1) is a solution on 5> = (-1, 1).
Note, however, that l/(x 2 - 1) is not defined at x = +1 and therefore could not be a solution on any interval
containing either of these two points.

1.7. Determine whether any of the functions (a) y l = sin 2x, (b) ;y 2(x) = x, or (c) y 3 (x) = -jsin2;t is a solution
to the initial-value problem y" + 4y = 0; y(0) = 0, /(O) = 1.
(a) y\(x) is a solution to the differential equation and satisfies the first initial condition y(0) = 0. However, y\(x)
does not satisfy the second initial condition (y[(x) = 2cos2;t;;yj(0) = 2cosO = 2^1); hence it is not a solution to the
initial-value problem, (b) y^(x) satisfies both initial conditions but does not satisfy the differential equation; hence
y 2(x) is not a solution, (c) y 3(x) satisfies the differential equation and both initial conditions; therefore, it is a solu-
tion to the initial-value problem.

1.8. Find the solution to the initial-value problem / + y = 0; y(3) = 2, if the general solution to the differential
equation is known to be (see Chapter 8) y(x) = c^'*, where c x is an arbitrary constant.
Since y(x) is a solution of the differential equation for every value of c 1; we seek that value of Cj which will
also satisfy the initial condition. Note that y(3) = c^e^. To satisfy the initial condition y(3) = 2, it is sufficient
3
3
to choose c l so that c^ = 2, that is, to choose c l = 2e . Substituting this value for c l into y(x), we obtain
3 x
x
3
y(x) = 2e e~ = 2e ^ as the solution of the initial-value problem.
1.9. Find a solution to the initial-value problem /' + 4y = 0; y(0) = 0, /(O) = 1, if the general solution to the
differential equation is known to be (see Chapter 9) y (x) = c1 sin 2x + c2 sos 2x.
Since y(x) is a solution of the differential equation for all values of Cj and c 2 (see Example 1.4), we seek those
values of Cj and c 2 that will also satisfy the initial conditions. Note that y(0) = Cj sin 0 + c 2 cos 0 = c 2. To satisfy
the first initial condition, y(0) = 0, we choose c 2 = 0. Furthermore, y'(x) = 2c l cos 2x - 2c 2 sin 2x; thus,